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Solutions Manual for Foundations Of MEMS 2nd Edition by Chang Liu

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The document contains multiple physics problems focusing on dynamics, including calculations for a sky diver's trajectory, projectile motion of an arrow, and drag forces on a sphere. Each problem provides equations of motion, integrates variables, and offers solutions with a focus on deriving parameters like speed and angle while considering drag and other forces. The problems also include dimensional analysis and unit conversions for various physical quantities.

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Problem 1.15 [Difficulty: 5] Given: Data on sky diver: M 70 kg⋅= kvert 0.25 N s 2 ⋅ m 2 ⋅= khoriz 0.05 N s 2 ⋅ m 2 ⋅= U0 70 m s ⋅= Find: Plot of trajectory. Solution: Use given data; integrate equation of motion by separating variables. Treat the sky diver as a system; apply Newton's 2nd law in horizontal and vertical directions: Vertical: Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): M dV dt ⋅ M g⋅ kvert V 2 ⋅−= (1) For V(t) we need to integrate (1) with respect to t: Separating variables and integrating: 0 V V V M g⋅ kvert V 2 − ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d 0 t t1 ⌠ ⎮ ⌡ d= so t 1 2 M kvert g⋅ ⋅ ln M g⋅ kvert V+ M g⋅ kvert V− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅= Rearranging o r V t( ) M g⋅ kvert e 2 kvert g⋅ M ⋅ t⋅ 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ e 2 kvert g⋅ M ⋅ t⋅ 1+ ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ ⋅= so V t( ) M g⋅ kvert tanh kvert g⋅ M t⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= For y(t) we need to integrate again: dy dt V= or y tV ⌠ ⎮ ⌡ d= y t( ) 0 t tV t( ) ⌠ ⎮ ⌡ d= 0 t t M g⋅ kvert tanh kvert g⋅ M t⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ ⌠ ⎮ ⎮ ⎮ ⌡ d= M g⋅ kvert ln cosh kvert g⋅ M t⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= y t( ) M g⋅ kvert ln cosh kvert g⋅ M t⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= Solutions Manual for Fox And Mcdonald's Introduction To Fluid Mechanics 8th Edition by Pritchard Full Download: https://downloadlink.org/p/solutions-manual-for-fox-and-mcdonalds-introduction-to-fluid-mechanics-8th-edition-b Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com
After the first few seconds we reach steady state: 0 1 2 3 4 5 10 20 30 t(s) y(m) y t( ) t 0 20 40 60 200 400 600 t(s) y(m) y t( ) t Horizontal: Newton's 2nd law for the sky diver (mass M) is: M dU dt ⋅ khoriz− U 2 ⋅= (2) For U(t) we need to integrate (2) with respect to t: Separating variables and integrating: U0 U U 1 U 2 ⌠ ⎮ ⎮ ⎮ ⌡ d 0 t t khoriz M − ⌠ ⎮ ⎮ ⌡ d= so khoriz M − t⋅ 1 U − 1 U0 += Rearranging U t( ) U0 1 khoriz U0⋅ M t⋅+ = For x(t) we need to integrate again: dx dt U= or x tU ⌠ ⎮ ⌡ d= x t( ) 0 t tU t( ) ⌠ ⎮ ⌡ d= 0 t t U0 1 khoriz U0⋅ M t⋅+ ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d= M khoriz ln khoriz U0⋅ M t⋅ 1+ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= x t( ) M khoriz ln khoriz U0⋅ M t⋅ 1+ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅=
0 20 40 60 0.5 1 1.5 2 t(s) x(km) Plotting the trajectory: 0 1 2 3 3− 2− 1− x(km) y(km) These plots can also be done in Excel.
Problem 1.16 [Difficulty: 3] Given: Long bow at range, R = 100 m. Maximum height of arrow is h = 10 m. Neglect air resistance. Find: Estimate of (a) speed, and (b) angle, of arrow leaving the bow. Plot: (a) release speed, and (b) angle, as a function of h Solution: Let V u i v j V i j)0 0 0= + = +0 0 0(cos sinθ θ ΣF m mgy dv dt = = − , so v = v0 – gt, and tf = 2tv=0 = 2v0/g Also, mv dv dy mg, v dv g dy, 0 v 2 gh0 2 = − = − − = − Thus h v 2g0 2 = (1) ΣF m du dt 0, so u u const, and R u t 2u v g 0 0 f 0 0 x = = = = = = (2) From Eq. 1: v 2gh0 2 = (3) From Eq. 2: u gR 2v gR 2 2gh u gR 8h 0 0 0 2 2 = = ∴ = Then 2 1 2 0 2 2 0 2 0 2 0 8 2and2 8 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +=+=+= h gR ghVgh h gR vuV (4) s m 7.37 m10 1 m100 s m 8 81.9 m10 s m 81.92 2 1 22 220 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ××+××=V From Eq. 3: v 2gh V sin sin 2gh V 0 0 1 0 = = = − θ θ, (5) R V0 θ0 y x h
°= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ××= − 8.21 m37.7 s m10 s m 81.92sin 2 1 2 1 θ Plots of V0 = V0(h) (Eq. 4) and θ0 = θ 0(h) (Eq. 5) are presented below: Initial Speed vs Maximum Height 0 10 20 30 40 50 60 70 80 0 5 10 15 20 25 30 h (m) V0(m/s) Initial Angle vs Maximum Height 0 10 20 30 40 50 60 0 5 10 15 20 25 30 h (m) θ(o )
Problem 1.17 [Difficulty: 2] Given: Basic dimensions M, L, t and T. Find: Dimensional representation of quantities below, and typical units in SI and English systems. Solution: (a) Power Power Energy Time Force Distance× Time == F L⋅ t = From Newton's 2nd law Force Mass Acceleration×= so F M L⋅ t 2 = Hence Power F L⋅ t = M L⋅ L⋅ t 2 t⋅ = M L 2 ⋅ t 3 = kg m 2 ⋅ s 3 slug ft 2 ⋅ s 3 (b) Pressure Pressure Force Area = F L 2 = M L⋅ t 2 L 2 ⋅ = M L t 2 ⋅ = kg m s 2 ⋅ slug ft s 2 ⋅ (c) Modulus of elasticity Pressure Force Area = F L 2 = M L⋅ t 2 L 2 ⋅ = M L t 2 ⋅ = kg m s 2 ⋅ slug ft s 2 ⋅ (d) Angular velocity AngularVelocity Radians Time = 1 t = 1 s 1 s (e) Energy Energy Force Distance×= F L⋅= M L⋅ L⋅ t 2 = M L 2 ⋅ t 2 = kg m 2 ⋅ s 2 slug ft 2 ⋅ s 2 (f) Moment of a force MomentOfForce Force Length×= F L⋅= M L⋅ L⋅ t 2 = M L 2 ⋅ t 2 = kg m 2 ⋅ s 2 slug ft 2 ⋅ s 2 (g) Momentum Momentum Mass Velocity×= M L t ⋅= M L⋅ t = kg m⋅ s slug ft⋅ s (h) Shear stress ShearStress Force Area = F L 2 = M L⋅ t 2 L 2 ⋅ = M L t 2 ⋅ = kg m s 2 ⋅ slug ft s 2 ⋅ (i) Strain Strain LengthChange Length = L L = Dimensionless (j) Angular momentum AngularMomentum Momentum Distance×= M L⋅ t L⋅= M L 2 ⋅ t = kg m 2 ⋅ s slugs ft 2 ⋅ s
Problem 1.18 [Difficulty: 2] Given: Basic dimensions F, L, t and T. Find: Dimensional representation of quantities below, and typical units in SI and English systems. Solution: (a) Power Power Energy Time Force Distance× Time == F L⋅ t = N m⋅ s lbf ft⋅ s (b) Pressure Pressure Force Area = F L 2 = N m 2 lbf ft 2 (c) Modulus of elasticity Pressure Force Area = F L 2 = N m 2 lbf ft 2 (d) Angular velocity AngularVelocity Radians Time = 1 t = 1 s 1 s (e) Energy Energy Force Distance×= F L⋅= N m⋅ lbf ft⋅ (f) Momentum Momentum Mass Velocity×= M L t ⋅= From Newton's 2nd law Force Mass Acceleration×= so F M L t 2 ⋅= or M F t 2 ⋅ L = Hence Momentum M L t ⋅= F t 2 ⋅ L⋅ L t⋅ = F t⋅= N s⋅ lbf s⋅ (g) Shear stress ShearStress Force Area = F L 2 = N m 2 lbf ft 2 (h) Specific heat SpecificHeat Energy Mass Temperature× = F L⋅ M T⋅ = F L⋅ F t 2 ⋅ L ⎛ ⎜ ⎝ ⎞ ⎠ T⋅ = L 2 t 2 T⋅ = m 2 s 2 K⋅ ft 2 s 2 R⋅ (i) Thermal expansion coefficient ThermalExpansionCoefficient LengthChange Length Temperature = 1 T = 1 K 1 R (j) Angular momentum AngularMomentum Momentum Distance×= F t⋅ L⋅= N m⋅ s⋅ lbf ft⋅ s⋅
Problem 1.19 [Difficulty: 1] Given: Viscosity, power, and specific energy data in certain units Find: Convert to different units Solution: Using data from tables (e.g. Table G.2) (a) 1 m 2 s ⋅ 1 m 2 s ⋅ 1 12 ft⋅ 0.0254 m⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ 2 ×= 10.76 ft 2 s ⋅= (b) 100 W⋅ 100 W⋅ 1 hp⋅ 746 W⋅ ×= 0.134 hp⋅= (c) 1 kJ kg ⋅ 1 kJ kg ⋅ 1000 J⋅ 1 kJ⋅ × 1 Btu⋅ 1055 J⋅ × 0.454 kg⋅ 1 lbm⋅ ×= 0.43 Btu lbm ⋅=
Problem 1.10 [Difficulty: 4] NOTE: Drag formula is in error: It should be: FD 3 π⋅ V⋅ d⋅= Mg FD = 3πVd a = dV/dt Given: Data on sphere and formula for drag. Find: Diameter of gasoline droplets that take 1 second to fall 10 in. Solution: Use given data and data in Appendices; integrate equation of motion by separating variables. The data provided, or available in the Appendices, are: μ 4.48 10 7− × lbf s⋅ ft 2 ⋅= ρw 1.94 slug ft 3 ⋅= SGgas 0.72= ρgas SGgas ρw⋅= ρgas 1.40 slug ft 3 ⋅= Newton's 2nd law for the sphere (mass M) is (ignoring buoyancy effects) M dV dt ⋅ M g⋅ 3 π⋅ μ⋅ V⋅ d⋅−= dV g 3 π⋅ μ⋅ d⋅ M V⋅− dt= so Integrating twice and using limits V t( ) M g⋅ 3 π⋅ μ⋅ d⋅ 1 e 3− π⋅ μ⋅ d⋅ M t⋅ − ⎛ ⎜ ⎝ ⎞ ⎠⋅= x t( ) M g⋅ 3 π⋅ μ⋅ d⋅ t M 3 π⋅ μ⋅ d⋅ e 3− π⋅ μ⋅ d⋅ M t⋅ 1− ⎛ ⎜ ⎝ ⎞ ⎠⋅+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ⋅= Replacing M with an expression involving diameter d M ρgas π d 3 ⋅ 6 ⋅= x t( ) ρgas d 2 ⋅ g⋅ 18 μ⋅ t ρgas d 2 ⋅ 18 μ⋅ e 18− μ⋅ ρgas d 2 ⋅ t⋅ 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠⋅+ ⎡⎢ ⎢ ⎢ ⎣ ⎤⎥ ⎥ ⎥ ⎦ ⋅= This equation must be solved for d so that x 1 s⋅( ) 10 in⋅= . The answer can be obtained from manual iteration, or by using Excel's Goal Seek. d 4.30 10 3− × in⋅= 0 0.025 0.05 0.075 0.1 0.25 0.5 0.75 1 t (s) x(in) 0 0.25 0.5 0.75 1 2.5 5 7.5 10 t (s) x(in) Note That the particle quickly reaches terminal speed, so that a simpler approximate solution would be to solve Mg = 3πµVd for d, with V = 0.25 m/s (allowing for the fact that M is a function of d)!
Problem 1.11 [Difficulty: 3] Given: Data on sphere and formula for drag. Find: Maximum speed, time to reach 95% of this speed, and plot speed as a function of time. Solution: Use given data and data in Appendices, and integrate equation of motion by separating variables. The data provided, or available in the Appendices, are: ρair 1.17 kg m 3 ⋅= μ 1.8 10 5− × N s⋅ m 2 ⋅= ρw 999 kg m 3 ⋅= SGSty 0.016= d 0.3 mm⋅= Then the density of the sphere is ρSty SGSty ρw⋅= ρSty 16 kg m 3 = The sphere mass is M ρSty π d 3 ⋅ 6 ⋅= 16 kg m 3 ⋅ π× 0.0003 m⋅( ) 3 6 ×= M 2.26 10 10− × kg= Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) M g⋅ 3 π⋅ V⋅ d⋅= so Vmax M g⋅ 3 π⋅ μ⋅ d⋅ = 1 3 π⋅ 2.26 10 10− ×× kg⋅ 9.81× m s 2 ⋅ m 2 1.8 10 5− × N⋅ s⋅ × 1 0.0003 m⋅ ×= Vmax 0.0435 m s = Newton's 2nd law for the general motion is (ignoring buoyancy effects) M dV dt ⋅ M g⋅ 3 π⋅ μ⋅ V⋅ d⋅−= Mg FD = 3πVd a = dV/dt so dV g 3 π⋅ μ⋅ d⋅ M V⋅− dt= Integrating and using limits V t( ) M g⋅ 3 π⋅ μ⋅ d⋅ 1 e 3− π⋅ μ⋅ d⋅ M t⋅ − ⎛ ⎜ ⎝ ⎞ ⎠⋅=
Using the given data 0 0.01 0.02 0.01 0.02 0.03 0.04 0.05 t (s) V(m/s) The time to reach 95% of maximum speed is obtained from M g⋅ 3 π⋅ μ⋅ d⋅ 1 e 3− π⋅ μ⋅ d⋅ M t⋅ − ⎛ ⎜ ⎝ ⎞ ⎠⋅ 0.95 Vmax⋅= so t M 3 π⋅ μ⋅ d⋅ − ln 1 0.95 Vmax⋅ 3⋅ π⋅ μ⋅ d⋅ M g⋅ − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= Substituting values t 0.0133 s= The plot can also be done in Excel.
Problem 1.12 [Difficulty: 3] mg kVt Given: Data on sphere and terminal speed. Find: Drag constant k, and time to reach 99% of terminal speed. Solution: Use given data; integrate equation of motion by separating variables. The data provided are: M 1 10 13− × slug⋅= Vt 0.2 ft s ⋅= Newton's 2nd law for the general motion is (ignoring buoyancy effects) M dV dt ⋅ M g⋅ k V⋅−= (1) Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) M g⋅ k Vt⋅= so k M g⋅ Vt = k 1 10 13− × slug⋅ 32.2× ft s 2 ⋅ s 0.2 ft⋅ × lbf s 2 ⋅ slug ft⋅ ×= k 1.61 10 11− × lbf s⋅ ft ⋅= dV g k M V⋅− dt= To find the time to reach 99% of Vt, we need V(t). From 1, separating variables Integrating and using limits t M k − ln 1 k M g⋅ V⋅−⎛ ⎜ ⎝ ⎞ ⎠ ⋅= We must evaluate this when V 0.99 Vt⋅= V 0.198 ft s ⋅= t 1− 10 13− × slug⋅ ft 1.61 10 11− × lbf⋅ s⋅ × lbf s 2 ⋅ slug ft⋅ × ln 1 1.61 10 11− × lbf s⋅ ft ⋅ 1 1 10 13− × slug⋅ × s 2 32.2 ft⋅ × 0.198 ft⋅ s × slug ft⋅ lbf s 2 ⋅ ×− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ ⋅= t 0.0286 s=
Problem 1.13 [Difficulty: 5] mg kVt Given: Data on sphere and terminal speed from Problem 1.12. Find: Distance traveled to reach 99% of terminal speed; plot of distance versus time. Solution: Use given data; integrate equation of motion by separating variables. The data provided are: M 1 10 13− × slug⋅= Vt 0.2 ft s ⋅= Newton's 2nd law for the general motion is (ignoring buoyancy effects) M dV dt ⋅ M g⋅ k V⋅−= (1) Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) M g⋅ k Vt⋅= so k M g⋅ Vt = k 1 10 13− × slug⋅ 32.2× ft s 2 ⋅ s 0.2 ft⋅ × lbf s 2 ⋅ slug ft⋅ ×= k 1.61 10 11− × lbf s⋅ ft ⋅= To find the distance to reach 99% of Vt, we need V(y). From 1: M dV dt ⋅ M dy dt ⋅ dV dy ⋅= M V⋅ dV dy ⋅= M g⋅ k V⋅−= V dV⋅ g k M V⋅− dy= Separating variables Integrating and using limits y M 2 g⋅ k 2 − ln 1 k M g⋅ V⋅−⎛ ⎜ ⎝ ⎞ ⎠ ⋅ M k V⋅−= We must evaluate this when V 0.99 Vt⋅= V 0.198 ft s ⋅= y 1 10 13− ⋅ slug⋅( ) 2 32.2 ft⋅ s 2 ⋅ ft 1.61 10 11− ⋅ lbf⋅ s⋅ ⎛ ⎜ ⎝ ⎞ ⎠ 2 ⋅ lbf s 2 ⋅ slug ft⋅ ⎛ ⎜ ⎝ ⎞ ⎠ 2 ⋅ ln 1 1.61 10 11− ⋅ lbf s⋅ ft ⋅ 1 1 10 13− ⋅ slug⋅ ⋅ s 2 32.2 ft⋅ ⋅ 0.198 ft⋅ s ⋅ slug ft⋅ lbf s 2 ⋅ ⋅− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ ⋅ 1 10 13− ⋅ slug⋅ ft 1.61 10 11− ⋅ lbf⋅ s⋅ × 0.198 ft⋅ s × lbf s 2 ⋅ slug ft⋅ ×+ ...= y 4.49 10 3− × ft⋅= Alternatively we could use the approach of Problem 1.12 and first find the time to reach terminal speed, and use this time in y(t) to find the above value of y: dV g k M V⋅− dt= From 1, separating variables Integrating and using limits t M k − ln 1 k M g⋅ V⋅−⎛ ⎜ ⎝ ⎞ ⎠ ⋅= (2)
We must evaluate this when V 0.99 Vt⋅= V 0.198 ft s ⋅= t 1 10 13− × slug⋅ ft 1.61 10 11− × lbf⋅ s⋅ × lbf s 2 ⋅ slug ft⋅ ⋅ ln 1 1.61 10 11− × lbf s⋅ ft ⋅ 1 1 10 13− × slug⋅ × s 2 32.2 ft⋅ × 0.198 ft⋅ s × slug ft⋅ lbf s 2 ⋅ ×− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ ⋅= t 0.0286 s= From 2, after rearranging V dy dt = M g⋅ k 1 e k M − t⋅ − ⎛ ⎜ ⎝ ⎞ ⎠⋅= Integrating and using limits y M g⋅ k t M k e k M − t⋅ 1− ⎛ ⎜ ⎝ ⎞ ⎠⋅+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ⋅= y 1 10 13− × slug⋅ 32.2 ft⋅ s 2 × ft 1.61 10 11− × lbf⋅ s⋅ × lbf s 2 ⋅ slug ft⋅ ⋅ 0.0291 s⋅ 10 13− slug⋅ ft 1.61 10 11− × lbf⋅ s⋅ ⋅ lbf s 2 ⋅ slug ft⋅ ⋅ e 1.61 10 11− × 1 10 13− × − .0291⋅ 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠⋅+ ...⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⋅= y 4.49 10 3− × ft⋅= 0 5 10 15 20 25 1.25 2.5 3.75 5 t (ms) y(0.001ft) This plot can also be presented in Excel.
Problem 1.14 [Difficulty: 4] Given: Data on sky diver: M 70 kg⋅= k 0.25 N s 2 ⋅ m 2 ⋅= Find: Maximum speed; speed after 100 m; plot speed as function of time and distance. Solution: Use given data; integrate equation of motion by separating variables. Treat the sky diver as a system; apply Newton's 2nd law: Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): M dV dt ⋅ M g⋅ k V 2 ⋅−= (1) Mg FD = kV2 a = dV/dt (a) For terminal speed Vt, acceleration is zero, so M g⋅ k V 2 ⋅− 0= so Vt M g⋅ k = Vt 70 kg⋅ 9.81× m s 2 ⋅ m 2 0.25 N⋅ s 2 ⋅ × N s 2 ⋅ kg m× ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ 1 2 = Vt 52.4 m s = (b) For V at y = 100 m we need to find V(y). From (1) M dV dt ⋅ M dV dy ⋅ dy dt ⋅= M V⋅ dV dt ⋅= M g⋅ k V 2 ⋅−= Separating variables and integrating: 0 V V V 1 k V 2 ⋅ M g⋅ − ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d 0 y yg ⌠ ⎮ ⌡ d= so ln 1 k V 2 ⋅ M g⋅ − ⎛ ⎜ ⎝ ⎞ ⎠ 2 k⋅ M − y= or V 2 M g⋅ k 1 e 2 k⋅ y⋅ M − − ⎛ ⎜ ⎝ ⎞ ⎠⋅= Hence V y( ) Vt 1 e 2 k⋅ y⋅ M − − ⎛ ⎜ ⎝ ⎞ ⎠ 1 2 ⋅= For y = 100 m: V 100 m⋅( ) 52.4 m s ⋅ 1 e 2− 0.25× N s 2 ⋅ m 2 ⋅ 100× m⋅ 1 70 kg⋅ × kg m⋅ s 2 N⋅ × − ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ 1 2 ⋅= V 100 m⋅( ) 37.4 m s ⋅=
0 100 200 300 400 500 20 40 60 y(m) V(m/s) (c) For V(t) we need to integrate (1) with respect to t: M dV dt ⋅ M g⋅ k V 2 ⋅−= Separating variables and integrating: 0 V V V M g⋅ k V 2 − ⌠ ⎮ ⎮ ⎮ ⌡ d 0 t t1 ⌠ ⎮ ⌡ d= so t 1 2 M k g⋅ ⋅ ln M g⋅ k V+ M g⋅ k V− ⎛⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅= 1 2 M k g⋅ ⋅ ln Vt V+ Vt V− ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= Rearranging V t( ) Vt e 2 k g⋅ M ⋅ t⋅ 1− ⎛ ⎜ ⎝ ⎞ ⎠ e 2 k g⋅ M ⋅ t⋅ 1+ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= or V t( ) Vt tanh Vt k M ⋅ t⋅⎛ ⎜ ⎝ ⎞ ⎠ ⋅= 0 5 10 15 20 20 40 60 t(s) V(m/s) V t( ) t The two graphs can also be plotted in Excel. Solutions Manual for Fox And Mcdonald's Introduction To Fluid Mechanics 8th Edition by Pritchard Full Download: https://downloadlink.org/p/solutions-manual-for-fox-and-mcdonalds-introduction-to-fluid-mechanics-8th-edition-by Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com

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Solutions Manual for Foundations Of MEMS 2nd Edition by Chang Liu

  • 1. Problem 1.15 [Difficulty: 5] Given: Data on sky diver: M 70 kg⋅= kvert 0.25 N s 2 ⋅ m 2 ⋅= khoriz 0.05 N s 2 ⋅ m 2 ⋅= U0 70 m s ⋅= Find: Plot of trajectory. Solution: Use given data; integrate equation of motion by separating variables. Treat the sky diver as a system; apply Newton's 2nd law in horizontal and vertical directions: Vertical: Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): M dV dt ⋅ M g⋅ kvert V 2 ⋅−= (1) For V(t) we need to integrate (1) with respect to t: Separating variables and integrating: 0 V V V M g⋅ kvert V 2 − ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d 0 t t1 ⌠ ⎮ ⌡ d= so t 1 2 M kvert g⋅ ⋅ ln M g⋅ kvert V+ M g⋅ kvert V− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅= Rearranging o r V t( ) M g⋅ kvert e 2 kvert g⋅ M ⋅ t⋅ 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ e 2 kvert g⋅ M ⋅ t⋅ 1+ ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ ⋅= so V t( ) M g⋅ kvert tanh kvert g⋅ M t⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= For y(t) we need to integrate again: dy dt V= or y tV ⌠ ⎮ ⌡ d= y t( ) 0 t tV t( ) ⌠ ⎮ ⌡ d= 0 t t M g⋅ kvert tanh kvert g⋅ M t⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ ⌠ ⎮ ⎮ ⎮ ⌡ d= M g⋅ kvert ln cosh kvert g⋅ M t⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= y t( ) M g⋅ kvert ln cosh kvert g⋅ M t⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= Solutions Manual for Fox And Mcdonald's Introduction To Fluid Mechanics 8th Edition by Pritchard Full Download: https://downloadlink.org/p/solutions-manual-for-fox-and-mcdonalds-introduction-to-fluid-mechanics-8th-edition-b Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com
  • 2. After the first few seconds we reach steady state: 0 1 2 3 4 5 10 20 30 t(s) y(m) y t( ) t 0 20 40 60 200 400 600 t(s) y(m) y t( ) t Horizontal: Newton's 2nd law for the sky diver (mass M) is: M dU dt ⋅ khoriz− U 2 ⋅= (2) For U(t) we need to integrate (2) with respect to t: Separating variables and integrating: U0 U U 1 U 2 ⌠ ⎮ ⎮ ⎮ ⌡ d 0 t t khoriz M − ⌠ ⎮ ⎮ ⌡ d= so khoriz M − t⋅ 1 U − 1 U0 += Rearranging U t( ) U0 1 khoriz U0⋅ M t⋅+ = For x(t) we need to integrate again: dx dt U= or x tU ⌠ ⎮ ⌡ d= x t( ) 0 t tU t( ) ⌠ ⎮ ⌡ d= 0 t t U0 1 khoriz U0⋅ M t⋅+ ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d= M khoriz ln khoriz U0⋅ M t⋅ 1+ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= x t( ) M khoriz ln khoriz U0⋅ M t⋅ 1+ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅=
  • 3. 0 20 40 60 0.5 1 1.5 2 t(s) x(km) Plotting the trajectory: 0 1 2 3 3− 2− 1− x(km) y(km) These plots can also be done in Excel.
  • 4. Problem 1.16 [Difficulty: 3] Given: Long bow at range, R = 100 m. Maximum height of arrow is h = 10 m. Neglect air resistance. Find: Estimate of (a) speed, and (b) angle, of arrow leaving the bow. Plot: (a) release speed, and (b) angle, as a function of h Solution: Let V u i v j V i j)0 0 0= + = +0 0 0(cos sinθ θ ΣF m mgy dv dt = = − , so v = v0 – gt, and tf = 2tv=0 = 2v0/g Also, mv dv dy mg, v dv g dy, 0 v 2 gh0 2 = − = − − = − Thus h v 2g0 2 = (1) ΣF m du dt 0, so u u const, and R u t 2u v g 0 0 f 0 0 x = = = = = = (2) From Eq. 1: v 2gh0 2 = (3) From Eq. 2: u gR 2v gR 2 2gh u gR 8h 0 0 0 2 2 = = ∴ = Then 2 1 2 0 2 2 0 2 0 2 0 8 2and2 8 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +=+=+= h gR ghVgh h gR vuV (4) s m 7.37 m10 1 m100 s m 8 81.9 m10 s m 81.92 2 1 22 220 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ××+××=V From Eq. 3: v 2gh V sin sin 2gh V 0 0 1 0 = = = − θ θ, (5) R V0 θ0 y x h
  • 5. °= ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ××= − 8.21 m37.7 s m10 s m 81.92sin 2 1 2 1 θ Plots of V0 = V0(h) (Eq. 4) and θ0 = θ 0(h) (Eq. 5) are presented below: Initial Speed vs Maximum Height 0 10 20 30 40 50 60 70 80 0 5 10 15 20 25 30 h (m) V0(m/s) Initial Angle vs Maximum Height 0 10 20 30 40 50 60 0 5 10 15 20 25 30 h (m) θ(o )
  • 6. Problem 1.17 [Difficulty: 2] Given: Basic dimensions M, L, t and T. Find: Dimensional representation of quantities below, and typical units in SI and English systems. Solution: (a) Power Power Energy Time Force Distance× Time == F L⋅ t = From Newton's 2nd law Force Mass Acceleration×= so F M L⋅ t 2 = Hence Power F L⋅ t = M L⋅ L⋅ t 2 t⋅ = M L 2 ⋅ t 3 = kg m 2 ⋅ s 3 slug ft 2 ⋅ s 3 (b) Pressure Pressure Force Area = F L 2 = M L⋅ t 2 L 2 ⋅ = M L t 2 ⋅ = kg m s 2 ⋅ slug ft s 2 ⋅ (c) Modulus of elasticity Pressure Force Area = F L 2 = M L⋅ t 2 L 2 ⋅ = M L t 2 ⋅ = kg m s 2 ⋅ slug ft s 2 ⋅ (d) Angular velocity AngularVelocity Radians Time = 1 t = 1 s 1 s (e) Energy Energy Force Distance×= F L⋅= M L⋅ L⋅ t 2 = M L 2 ⋅ t 2 = kg m 2 ⋅ s 2 slug ft 2 ⋅ s 2 (f) Moment of a force MomentOfForce Force Length×= F L⋅= M L⋅ L⋅ t 2 = M L 2 ⋅ t 2 = kg m 2 ⋅ s 2 slug ft 2 ⋅ s 2 (g) Momentum Momentum Mass Velocity×= M L t ⋅= M L⋅ t = kg m⋅ s slug ft⋅ s (h) Shear stress ShearStress Force Area = F L 2 = M L⋅ t 2 L 2 ⋅ = M L t 2 ⋅ = kg m s 2 ⋅ slug ft s 2 ⋅ (i) Strain Strain LengthChange Length = L L = Dimensionless (j) Angular momentum AngularMomentum Momentum Distance×= M L⋅ t L⋅= M L 2 ⋅ t = kg m 2 ⋅ s slugs ft 2 ⋅ s
  • 7. Problem 1.18 [Difficulty: 2] Given: Basic dimensions F, L, t and T. Find: Dimensional representation of quantities below, and typical units in SI and English systems. Solution: (a) Power Power Energy Time Force Distance× Time == F L⋅ t = N m⋅ s lbf ft⋅ s (b) Pressure Pressure Force Area = F L 2 = N m 2 lbf ft 2 (c) Modulus of elasticity Pressure Force Area = F L 2 = N m 2 lbf ft 2 (d) Angular velocity AngularVelocity Radians Time = 1 t = 1 s 1 s (e) Energy Energy Force Distance×= F L⋅= N m⋅ lbf ft⋅ (f) Momentum Momentum Mass Velocity×= M L t ⋅= From Newton's 2nd law Force Mass Acceleration×= so F M L t 2 ⋅= or M F t 2 ⋅ L = Hence Momentum M L t ⋅= F t 2 ⋅ L⋅ L t⋅ = F t⋅= N s⋅ lbf s⋅ (g) Shear stress ShearStress Force Area = F L 2 = N m 2 lbf ft 2 (h) Specific heat SpecificHeat Energy Mass Temperature× = F L⋅ M T⋅ = F L⋅ F t 2 ⋅ L ⎛ ⎜ ⎝ ⎞ ⎠ T⋅ = L 2 t 2 T⋅ = m 2 s 2 K⋅ ft 2 s 2 R⋅ (i) Thermal expansion coefficient ThermalExpansionCoefficient LengthChange Length Temperature = 1 T = 1 K 1 R (j) Angular momentum AngularMomentum Momentum Distance×= F t⋅ L⋅= N m⋅ s⋅ lbf ft⋅ s⋅
  • 8. Problem 1.19 [Difficulty: 1] Given: Viscosity, power, and specific energy data in certain units Find: Convert to different units Solution: Using data from tables (e.g. Table G.2) (a) 1 m 2 s ⋅ 1 m 2 s ⋅ 1 12 ft⋅ 0.0254 m⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ 2 ×= 10.76 ft 2 s ⋅= (b) 100 W⋅ 100 W⋅ 1 hp⋅ 746 W⋅ ×= 0.134 hp⋅= (c) 1 kJ kg ⋅ 1 kJ kg ⋅ 1000 J⋅ 1 kJ⋅ × 1 Btu⋅ 1055 J⋅ × 0.454 kg⋅ 1 lbm⋅ ×= 0.43 Btu lbm ⋅=
  • 9. Problem 1.10 [Difficulty: 4] NOTE: Drag formula is in error: It should be: FD 3 π⋅ V⋅ d⋅= Mg FD = 3πVd a = dV/dt Given: Data on sphere and formula for drag. Find: Diameter of gasoline droplets that take 1 second to fall 10 in. Solution: Use given data and data in Appendices; integrate equation of motion by separating variables. The data provided, or available in the Appendices, are: μ 4.48 10 7− × lbf s⋅ ft 2 ⋅= ρw 1.94 slug ft 3 ⋅= SGgas 0.72= ρgas SGgas ρw⋅= ρgas 1.40 slug ft 3 ⋅= Newton's 2nd law for the sphere (mass M) is (ignoring buoyancy effects) M dV dt ⋅ M g⋅ 3 π⋅ μ⋅ V⋅ d⋅−= dV g 3 π⋅ μ⋅ d⋅ M V⋅− dt= so Integrating twice and using limits V t( ) M g⋅ 3 π⋅ μ⋅ d⋅ 1 e 3− π⋅ μ⋅ d⋅ M t⋅ − ⎛ ⎜ ⎝ ⎞ ⎠⋅= x t( ) M g⋅ 3 π⋅ μ⋅ d⋅ t M 3 π⋅ μ⋅ d⋅ e 3− π⋅ μ⋅ d⋅ M t⋅ 1− ⎛ ⎜ ⎝ ⎞ ⎠⋅+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ⋅= Replacing M with an expression involving diameter d M ρgas π d 3 ⋅ 6 ⋅= x t( ) ρgas d 2 ⋅ g⋅ 18 μ⋅ t ρgas d 2 ⋅ 18 μ⋅ e 18− μ⋅ ρgas d 2 ⋅ t⋅ 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠⋅+ ⎡⎢ ⎢ ⎢ ⎣ ⎤⎥ ⎥ ⎥ ⎦ ⋅= This equation must be solved for d so that x 1 s⋅( ) 10 in⋅= . The answer can be obtained from manual iteration, or by using Excel's Goal Seek. d 4.30 10 3− × in⋅= 0 0.025 0.05 0.075 0.1 0.25 0.5 0.75 1 t (s) x(in) 0 0.25 0.5 0.75 1 2.5 5 7.5 10 t (s) x(in) Note That the particle quickly reaches terminal speed, so that a simpler approximate solution would be to solve Mg = 3πµVd for d, with V = 0.25 m/s (allowing for the fact that M is a function of d)!
  • 10. Problem 1.11 [Difficulty: 3] Given: Data on sphere and formula for drag. Find: Maximum speed, time to reach 95% of this speed, and plot speed as a function of time. Solution: Use given data and data in Appendices, and integrate equation of motion by separating variables. The data provided, or available in the Appendices, are: ρair 1.17 kg m 3 ⋅= μ 1.8 10 5− × N s⋅ m 2 ⋅= ρw 999 kg m 3 ⋅= SGSty 0.016= d 0.3 mm⋅= Then the density of the sphere is ρSty SGSty ρw⋅= ρSty 16 kg m 3 = The sphere mass is M ρSty π d 3 ⋅ 6 ⋅= 16 kg m 3 ⋅ π× 0.0003 m⋅( ) 3 6 ×= M 2.26 10 10− × kg= Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) M g⋅ 3 π⋅ V⋅ d⋅= so Vmax M g⋅ 3 π⋅ μ⋅ d⋅ = 1 3 π⋅ 2.26 10 10− ×× kg⋅ 9.81× m s 2 ⋅ m 2 1.8 10 5− × N⋅ s⋅ × 1 0.0003 m⋅ ×= Vmax 0.0435 m s = Newton's 2nd law for the general motion is (ignoring buoyancy effects) M dV dt ⋅ M g⋅ 3 π⋅ μ⋅ V⋅ d⋅−= Mg FD = 3πVd a = dV/dt so dV g 3 π⋅ μ⋅ d⋅ M V⋅− dt= Integrating and using limits V t( ) M g⋅ 3 π⋅ μ⋅ d⋅ 1 e 3− π⋅ μ⋅ d⋅ M t⋅ − ⎛ ⎜ ⎝ ⎞ ⎠⋅=
  • 11. Using the given data 0 0.01 0.02 0.01 0.02 0.03 0.04 0.05 t (s) V(m/s) The time to reach 95% of maximum speed is obtained from M g⋅ 3 π⋅ μ⋅ d⋅ 1 e 3− π⋅ μ⋅ d⋅ M t⋅ − ⎛ ⎜ ⎝ ⎞ ⎠⋅ 0.95 Vmax⋅= so t M 3 π⋅ μ⋅ d⋅ − ln 1 0.95 Vmax⋅ 3⋅ π⋅ μ⋅ d⋅ M g⋅ − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= Substituting values t 0.0133 s= The plot can also be done in Excel.
  • 12. Problem 1.12 [Difficulty: 3] mg kVt Given: Data on sphere and terminal speed. Find: Drag constant k, and time to reach 99% of terminal speed. Solution: Use given data; integrate equation of motion by separating variables. The data provided are: M 1 10 13− × slug⋅= Vt 0.2 ft s ⋅= Newton's 2nd law for the general motion is (ignoring buoyancy effects) M dV dt ⋅ M g⋅ k V⋅−= (1) Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) M g⋅ k Vt⋅= so k M g⋅ Vt = k 1 10 13− × slug⋅ 32.2× ft s 2 ⋅ s 0.2 ft⋅ × lbf s 2 ⋅ slug ft⋅ ×= k 1.61 10 11− × lbf s⋅ ft ⋅= dV g k M V⋅− dt= To find the time to reach 99% of Vt, we need V(t). From 1, separating variables Integrating and using limits t M k − ln 1 k M g⋅ V⋅−⎛ ⎜ ⎝ ⎞ ⎠ ⋅= We must evaluate this when V 0.99 Vt⋅= V 0.198 ft s ⋅= t 1− 10 13− × slug⋅ ft 1.61 10 11− × lbf⋅ s⋅ × lbf s 2 ⋅ slug ft⋅ × ln 1 1.61 10 11− × lbf s⋅ ft ⋅ 1 1 10 13− × slug⋅ × s 2 32.2 ft⋅ × 0.198 ft⋅ s × slug ft⋅ lbf s 2 ⋅ ×− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ ⋅= t 0.0286 s=
  • 13. Problem 1.13 [Difficulty: 5] mg kVt Given: Data on sphere and terminal speed from Problem 1.12. Find: Distance traveled to reach 99% of terminal speed; plot of distance versus time. Solution: Use given data; integrate equation of motion by separating variables. The data provided are: M 1 10 13− × slug⋅= Vt 0.2 ft s ⋅= Newton's 2nd law for the general motion is (ignoring buoyancy effects) M dV dt ⋅ M g⋅ k V⋅−= (1) Newton's 2nd law for the steady state motion becomes (ignoring buoyancy effects) M g⋅ k Vt⋅= so k M g⋅ Vt = k 1 10 13− × slug⋅ 32.2× ft s 2 ⋅ s 0.2 ft⋅ × lbf s 2 ⋅ slug ft⋅ ×= k 1.61 10 11− × lbf s⋅ ft ⋅= To find the distance to reach 99% of Vt, we need V(y). From 1: M dV dt ⋅ M dy dt ⋅ dV dy ⋅= M V⋅ dV dy ⋅= M g⋅ k V⋅−= V dV⋅ g k M V⋅− dy= Separating variables Integrating and using limits y M 2 g⋅ k 2 − ln 1 k M g⋅ V⋅−⎛ ⎜ ⎝ ⎞ ⎠ ⋅ M k V⋅−= We must evaluate this when V 0.99 Vt⋅= V 0.198 ft s ⋅= y 1 10 13− ⋅ slug⋅( ) 2 32.2 ft⋅ s 2 ⋅ ft 1.61 10 11− ⋅ lbf⋅ s⋅ ⎛ ⎜ ⎝ ⎞ ⎠ 2 ⋅ lbf s 2 ⋅ slug ft⋅ ⎛ ⎜ ⎝ ⎞ ⎠ 2 ⋅ ln 1 1.61 10 11− ⋅ lbf s⋅ ft ⋅ 1 1 10 13− ⋅ slug⋅ ⋅ s 2 32.2 ft⋅ ⋅ 0.198 ft⋅ s ⋅ slug ft⋅ lbf s 2 ⋅ ⋅− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ ⋅ 1 10 13− ⋅ slug⋅ ft 1.61 10 11− ⋅ lbf⋅ s⋅ × 0.198 ft⋅ s × lbf s 2 ⋅ slug ft⋅ ×+ ...= y 4.49 10 3− × ft⋅= Alternatively we could use the approach of Problem 1.12 and first find the time to reach terminal speed, and use this time in y(t) to find the above value of y: dV g k M V⋅− dt= From 1, separating variables Integrating and using limits t M k − ln 1 k M g⋅ V⋅−⎛ ⎜ ⎝ ⎞ ⎠ ⋅= (2)
  • 14. We must evaluate this when V 0.99 Vt⋅= V 0.198 ft s ⋅= t 1 10 13− × slug⋅ ft 1.61 10 11− × lbf⋅ s⋅ × lbf s 2 ⋅ slug ft⋅ ⋅ ln 1 1.61 10 11− × lbf s⋅ ft ⋅ 1 1 10 13− × slug⋅ × s 2 32.2 ft⋅ × 0.198 ft⋅ s × slug ft⋅ lbf s 2 ⋅ ×− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ ⋅= t 0.0286 s= From 2, after rearranging V dy dt = M g⋅ k 1 e k M − t⋅ − ⎛ ⎜ ⎝ ⎞ ⎠⋅= Integrating and using limits y M g⋅ k t M k e k M − t⋅ 1− ⎛ ⎜ ⎝ ⎞ ⎠⋅+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ⋅= y 1 10 13− × slug⋅ 32.2 ft⋅ s 2 × ft 1.61 10 11− × lbf⋅ s⋅ × lbf s 2 ⋅ slug ft⋅ ⋅ 0.0291 s⋅ 10 13− slug⋅ ft 1.61 10 11− × lbf⋅ s⋅ ⋅ lbf s 2 ⋅ slug ft⋅ ⋅ e 1.61 10 11− × 1 10 13− × − .0291⋅ 1− ⎛ ⎜ ⎜ ⎝ ⎞ ⎠⋅+ ...⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⋅= y 4.49 10 3− × ft⋅= 0 5 10 15 20 25 1.25 2.5 3.75 5 t (ms) y(0.001ft) This plot can also be presented in Excel.
  • 15. Problem 1.14 [Difficulty: 4] Given: Data on sky diver: M 70 kg⋅= k 0.25 N s 2 ⋅ m 2 ⋅= Find: Maximum speed; speed after 100 m; plot speed as function of time and distance. Solution: Use given data; integrate equation of motion by separating variables. Treat the sky diver as a system; apply Newton's 2nd law: Newton's 2nd law for the sky diver (mass M) is (ignoring buoyancy effects): M dV dt ⋅ M g⋅ k V 2 ⋅−= (1) Mg FD = kV2 a = dV/dt (a) For terminal speed Vt, acceleration is zero, so M g⋅ k V 2 ⋅− 0= so Vt M g⋅ k = Vt 70 kg⋅ 9.81× m s 2 ⋅ m 2 0.25 N⋅ s 2 ⋅ × N s 2 ⋅ kg m× ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ 1 2 = Vt 52.4 m s = (b) For V at y = 100 m we need to find V(y). From (1) M dV dt ⋅ M dV dy ⋅ dy dt ⋅= M V⋅ dV dt ⋅= M g⋅ k V 2 ⋅−= Separating variables and integrating: 0 V V V 1 k V 2 ⋅ M g⋅ − ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d 0 y yg ⌠ ⎮ ⌡ d= so ln 1 k V 2 ⋅ M g⋅ − ⎛ ⎜ ⎝ ⎞ ⎠ 2 k⋅ M − y= or V 2 M g⋅ k 1 e 2 k⋅ y⋅ M − − ⎛ ⎜ ⎝ ⎞ ⎠⋅= Hence V y( ) Vt 1 e 2 k⋅ y⋅ M − − ⎛ ⎜ ⎝ ⎞ ⎠ 1 2 ⋅= For y = 100 m: V 100 m⋅( ) 52.4 m s ⋅ 1 e 2− 0.25× N s 2 ⋅ m 2 ⋅ 100× m⋅ 1 70 kg⋅ × kg m⋅ s 2 N⋅ × − ⎛ ⎜ ⎜ ⎝ ⎞ ⎠ 1 2 ⋅= V 100 m⋅( ) 37.4 m s ⋅=
  • 16. 0 100 200 300 400 500 20 40 60 y(m) V(m/s) (c) For V(t) we need to integrate (1) with respect to t: M dV dt ⋅ M g⋅ k V 2 ⋅−= Separating variables and integrating: 0 V V V M g⋅ k V 2 − ⌠ ⎮ ⎮ ⎮ ⌡ d 0 t t1 ⌠ ⎮ ⌡ d= so t 1 2 M k g⋅ ⋅ ln M g⋅ k V+ M g⋅ k V− ⎛⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅= 1 2 M k g⋅ ⋅ ln Vt V+ Vt V− ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= Rearranging V t( ) Vt e 2 k g⋅ M ⋅ t⋅ 1− ⎛ ⎜ ⎝ ⎞ ⎠ e 2 k g⋅ M ⋅ t⋅ 1+ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅= or V t( ) Vt tanh Vt k M ⋅ t⋅⎛ ⎜ ⎝ ⎞ ⎠ ⋅= 0 5 10 15 20 20 40 60 t(s) V(m/s) V t( ) t The two graphs can also be plotted in Excel. Solutions Manual for Fox And Mcdonald's Introduction To Fluid Mechanics 8th Edition by Pritchard Full Download: https://downloadlink.org/p/solutions-manual-for-fox-and-mcdonalds-introduction-to-fluid-mechanics-8th-edition-by Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com